A perfect gas goes from state A to state B by absorbing 8×105 J of heat and doing 6.5×105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process

Question

A perfect gas goes from state A to state B by absorbing 8×105 J of heat and doing 6.5×105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In the second process (A) work done on gas is 105 J (B) work done on gas is 0.5×105 J (C) work done by gas is 105 J (D) work done by gas is 0.5×105 J

Tardigrade Admin · Accepted Answer

dU = dQ - dW =(8 × 105-6.5 × 105)=1.5 × 105 J dW = dQ - dU =105-1.5 × 105=-0.5 × 105 J - ve sign indicates that work done on the gas is 0.5 × 105 J

Q. A perfect gas goes from state A to state B by absorbing $8 \times 1 0^{5} J$ of heat and doing $6.5 \times 1 0^{5} J$ of external work. It is now transferred between the same two states in another process in which it absorbs $1 0^{5} J$ of heat. In the second process

work done on gas is $1 0^{5} J$

work done on gas is $0.5 \times 1 0^{5} J$

work done by gas is $1 0^{5} J$

work done by gas is $0.5 \times 1 0^{5} J$

$d U = d Q - d W = (8 \times 1 0^{5} - 6.5 \times 1 0^{5}) = 1.5 \times 1 0^{5} J$
$d W = d Q - d U = 1 0^{5} - 1.5 \times 1 0^{5} = - 0.5 \times 1 0^{5} J$
- ve sign indicates that work done on the gas is $0.5 \times 1 0^{5} J$

Q. A perfect gas goes from state A to state B by absorbing 8×105J of heat and doing 6.5×105J of external work. It is now transferred between the same two states in another process in which it absorbs 105J of heat. In the second process

work done on gas is 105J

work done on gas is 0.5×105J

work done by gas is 105J

work done by gas is 0.5×105J