During an adiabatic compression, 830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50 %. The change in its temperature is nearly: (R = 8.3 JK-1 mol-1)

Question

During an adiabatic compression, 830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50 %. The change in its temperature is nearly: (R = 8.3 JK-1 mol-1) (A) 40 K (B) 33 K (C) 20 K (D) 14 K

Tardigrade Admin · Accepted Answer

The work done in adiabatic process is given as W=(n R Δ T/&gamma;-1) or 830=(2 × 8.3 × Δ T/1.4-1) Δ T=20 K

Q. During an adiabatic compression, $830 J$ of work is done on $2$ moles of a diatomic ideal gas to reduce its volume by $50%$ . The change in its temperature is nearly : $(R = 8.3 J K^{- 1} m o l^{- 1})$

40 K

33 K

20 K

14 K

The work done in adiabatic process is given as
$< b r / > W = \frac{n R Δ T}{γ - 1} < b r / >$
or
$830 = \frac{2 \times 8.3 \times Δ T}{1.4 - 1}$
$< b r / > Δ T = 20 K < b r / >$

Q. During an adiabatic compression, 830J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly : (R=8.3JK−1mol−1)