One mole of an ideal gas (CV = 20 J K- 1 m o l- 1) initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be-

Question

One mole of an ideal gas (CV = 20 J K- 1 m o l- 1) initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be- (A) W=0;q=5.46kJ (B) W=0;q=0 (C) W=-5.46kJ;q=5.46kJ (D) W=5.46kJ;q=5.46kJ

Tardigrade Admin · Accepted Answer

T1=273K T2=2× 273K ⇒ Δ T=273 Volume is constant So, Δ V=0 W=PΔ V=P× 0=0 q=nCV(Δ T)=20× 273=5460J=5.46kJ

Q. One mole of an ideal gas $(C_{V} = 20 J K^{- 1} m o l^{- 1})$ initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be-

$W = 0; q = 5.46 k J$

$W = 0; q = 0$

$W = - 5.46 k J; q = 5.46 k J$

$W = 5.46 k J; q = 5.46 k J$

$T_{1} = 273 K$

$T_{2} = 2 \times 273 K$

$\Rightarrow Δ T = 273$

Volume is constant

So, $Δ V = 0$

$W = P Δ V = P \times 0 = 0$

$q = n C_{V} (Δ T) = 20 \times 273 = 5460 J = 5.46 k J$

Q. One mole of an ideal gas (CV​=20JK−1mol−1) initially at STP is heated at constant volume to twice the initial temperature. For the process, W and q will be-

W=0;q=5.46kJ

W=0;q=0

W=−5.46kJ;q=5.46kJ

W=5.46kJ;q=5.46kJ